Remove duplicates from an array of small primes

Given an array of primes arr[] such that the range of primes is small. Remove duplicates from the array.

Examples:

Input: arr[] = {3, 5, 7, 2, 2, 5, 7, 7}
Output: 2 3 5 7
Explanation: There are no duplicates among {2, 3, 5, 7}.

Input: arr[] = {3, 5, 7, 3, 3, 13, 5, 13, 29, 13}
Output: 3 5 7 13 29
Explanation: There are no duplicates among {3, 5, 7, 13, 29}

Source: Amazon Interview Question

Approach: This method discusses the naive approach which takes O(N²) time complexity.

So, the basic idea is to check for every element, whether it has occurred previously or not. Therefore, the approach involves keeping two loops, one to select the present element or index and the inner loop to check if the element has previously occurred or not.

Step-by-step algorithm:

• Start by running two loops.
• Pick all elements one by one.
• For every picked element, check if it has already occurred or not.
• If already seen, then ignore it, else add it to the array.

Implementation:

C++

// A C++ program to implement Naive
// approach to remove duplicates.
#include <bits/stdc++.h>
using namespace std;

int removeDups(vector<int>& vect)
{
    int res_ind = 1;

    // Loop invariant: Elements from vect[0]
    // to vect[res_ind-1] are unique.
    for (int i = 1; i < vect.size(); i++) {
        int j;
        for (j = 0; j < i; j++)
            if (vect[i] == vect[j])
                break;
        if (j == i)
            vect[res_ind++] = vect[i];
    }

    // Removes elements from vect[res_ind] to
    // vect[end]
    vect.erase(vect.begin() + res_ind, vect.end());
}

// Driver code
int main()
{
    vector<int> vect{ 3, 5, 7, 2, 2, 5, 7, 7 };
    removeDups(vect);
    for (int i = 0; i < vect.size(); i++)
        cout << vect[i] << " ";
    return 0;
}

Java

// Java program to implement Naive
// approach to remove duplicates
class GFG {
    static int[] removeDups(int[] vect)
    {
        int res_ind = 1;

        // Loop invariant: Elements from vect[0]
        // to vect[res_ind-1] are unique.
        for (int i = 1; i < vect.length; i++) {
            int j;
            for (j = 0; j < i; j++)
                if (vect[i] == vect[j])
                    break;
            if (j == i)
                vect[res_ind++] = vect[i];
        }

        // Removes elements from vect[res_ind]
        // to vect[end]
        int[] new_arr = new int[res_ind];
        for (int i = 0; i < res_ind; i++)
            new_arr[i] = vect[i];

        return new_arr;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] vect = { 3, 5, 7, 2, 2, 5, 7, 7 };
        vect = removeDups(vect);

        for (int i = 0; i < vect.length; i++)
            System.out.print(vect[i] + " ");
        System.out.println();
    }
}

// This code is contributed by
// sanjeev2552

C#

// C# program to implement Naive approach
// to remove duplicates
using System;

class GFG {
    static int[] removeDups(int[] vect)
    {
        int res_ind = 1;

        // Loop invariant : Elements from vect[0]
        // to vect[res_ind-1] are unique.
        for (int i = 1; i < vect.Length; i++) {
            int j;
            for (j = 0; j < i; j++)
                if (vect[i] == vect[j])
                    break;
            if (j == i)
                vect[res_ind++] = vect[i];
        }

        // Removes elements from vect[res_ind]
        // to vect[end]
        int[] new_arr = new int[res_ind];
        for (int i = 0; i < res_ind; i++)
            new_arr[i] = vect[i];

        return new_arr;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int[] vect = { 3, 5, 7, 2, 2, 5, 7, 7 };
        vect = removeDups(vect);

        for (int i = 0; i < vect.Length; i++)
            Console.Write(vect[i] + " ");
        Console.WriteLine();
    }
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript program to implement Naive
// approach to remove duplicates

    function removeDups(vect)
    {
        let res_ind = 1;

        // Loop invariant: Elements from vect[0]
        // to vect[res_ind-1] are unique.
        for (let i = 1; i < vect.length; i++) {
            let j;
            for (j = 0; j < i; j++)
                if (vect[i] == vect[j])
                    break;
            if (j == i)
                vect[res_ind++] = vect[i];
        }

        // Removes elements from vect[res_ind]
        // to vect[end]
        let new_arr = new Array(res_ind);
        for (let i = 0; i < res_ind; i++)
            new_arr[i] = vect[i];

        return new_arr;
    }

      // Driver Code
    let vect=[3, 5, 7, 2, 2, 5, 7, 7];
    vect = removeDups(vect);
    for (let i = 0; i < vect.length; i++)
        document.write(vect[i] + " ");
    document.write("<br>");


// This code is contributed by unknown2108

</script>

Python3

# A Python3 program to implement
# Naive approach to remove duplicates.
def removeDups(vect):

    res_ind = 1

# Loop invariant : Elements from vect[0]
# to vect[res_ind-1] are unique.
    for i in range(1, len(vect)):
        j = 0
        while (j < i):
            if (vect[i] == vect[j]):
                break
            j += 1
        if (j == i):
            vect[res_ind] = vect[i]
            res_ind += 1

# Removes elements from
# vect[res_ind] to vect[end]
    return vect[0:res_ind]

# Driver code
vect = [3, 5, 7, 2, 2, 5, 7, 7]
vect = removeDups(vect)
for i in range(len(vect)):
    print(vect[i], end = " ")

# This code is contributed
# by mohit kumar

Output

3 5 7 2 

Complexity Analysis:

• Time Complexity: O(n²). 
  As two nested loops are used, so the time complexity becomes O(n²).
• Auxiliary Space: O(1).

Method 2: This method involves the technique of Sorting which takes O(n log n) time.

Approach:

In comparison to the previous approach, a better solution is to first sort the array and then remove all the adjacent elements which are similar, from the sorted array.

Step-by-step algorithm:

• First sort the array.
• The need for extra space can be cleverly avoided. Keeping two variables, first = 1 and i = 1.
• Traverse the array from the second element to the end.
• For every element, if that element is not equal to the previous element, then array[first++] = array[i], where i is the counter of a loop.
• So the length of the array with no duplicates is first, remove the rest elements.

Note: In CPP there are a few inbuilt functions like sort() to sort and unique() to remove adjacent duplicates. The unique() function puts all unique elements at the beginning and returns an iterator pointing to the first element after the unique element. The erase() function removes elements between two given iterators.

Implementation:

C++

// C++ program to remove duplicates using Sorting
#include <bits/stdc++.h>
using namespace std;

int removeDups(vector<int>& vect)
{
    // Sort the vector
    sort(vect.begin(), vect.end());

    // unique() removes adjacent duplicates.
    // unique function puts all unique elements at
    // the beginning and returns iterator pointing
    // to the first element after unique element.
    // Erase function removes elements between two
    // given iterators
    vect.erase(unique(vect.begin(), vect.end()),
               vect.end());
}

// Driver code
int main()
{
    vector<int> vect{ 3, 5, 7, 2, 2, 5, 7, 7 };
    removeDups(vect);
    for (int i = 0; i < vect.size(); i++)
        cout << vect[i] << " ";
    return 0;
}

Java

// Java program to remove duplicates using Sorting
import java.util.*;

class GFG {

    static int[] removeDups(int[] vect)
    {
        // sort the array
        Arrays.sort(vect);

        // pointer
        int first = 1;

        // remove duplicate elements
        for (int i = 1; i < vect.length; i++)
            if (vect[i] != vect[i - 1])
                vect[first++] = vect[i];

        // mark rest of elements to INT_MAX
        for (int i = first; i < vect.length; i++)
            vect[i] = Integer.MAX_VALUE;

        return vect;
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] vect = { 3, 5, 7, 2, 2, 5, 7, 7 };
        vect = removeDups(vect);
        for (int i = 0; i < vect.length; i++) {
            if (vect[i] == Integer.MAX_VALUE)
                break;
            System.out.print(vect[i] + " ");
        }
    }
}

C#

// C# program to remove duplicates using Sorting
using System;
class GFG
{
    static int[] removeDups(int[] vect)
    {

        // sort the array
        Array.Sort(vect);

        // pointer
        int first = 1;

        // remove duplicate elements
        for (int i = 1; i < vect.Length; i++)
            if (vect[i] != vect[i - 1])
                vect[first++] = vect[i];

        // mark rest of elements to INT_MAX
        for (int i = first; i < vect.Length; i++)
            vect[i] = int.MaxValue;
        return vect;
    }

    // Driver code
    public static void Main(params string[] args)
    {
        int[] vect = { 3, 5, 7, 2, 2, 5, 7, 7 };
        vect = removeDups(vect);
        for (int i = 0; i < vect.Length; i++)
        {
            if (vect[i] == int.MaxValue)
                break;
            Console.Write(vect[i] + " ");
        }
    }
}

// This code is contributed by rutvik_56.

Javascript

<script>
// Javascript program to remove duplicates using Sorting

function removeDups(vect) {
    // Sort the vector
    vect.sort((a, b) => a - b)

    // pointer
    let first = 1;

    // remove duplicate elements
    for (let i = 1; i < vect.length; i++){
        if (vect[i] != vect[i - 1])
            vect[first++] = vect[i];
    }

    // mark rest of elements to INT_MAX
    for (let i = first; i < vect.length; i++)
        vect[i] = Number.MAX_SAFE_INTEGER;

    return vect;
}

    // Driver code
    let vect = [ 3, 5, 7, 2, 2, 5, 7, 7 ];
    removeDups(vect);
    for (let i = 0; i < vect.length; i++) {
        if (vect[i] == Number.MAX_SAFE_INTEGER)
            break;
        document.write(vect[i] + " ");
    }

</script>

Python3

# Python3 program to remove duplicates
# using Sorting
import sys

def removeDups(vect):

    # Sort the array
    vect.sort()

    # Pointer
    first = 1

    # Remove duplicate elements
    for i in range(1, len(vect)):
        if (vect[i] != vect[i - 1]):
            vect[first] = vect[i]
            first += 1

    # Mark rest of elements to INT_MAX
    for  i in range(first, len(vect)):
        vect[i] = sys.maxsize

    return vect

# Driver code
vect = [ 3, 5, 7, 2, 2, 5, 7, 7 ]
vect = removeDups(vect)

for i in range(len(vect)):
    if (vect[i] == sys.maxsize):
        break

    print(vect[i], end = " ")

# This code is contributed by avanitrachhadiya2155

Output

2 3 5 7 

Output:

Complexity Analysis:

• Time Complexity: O(n Log n). 
  For sorting the array O(n log n ) time complexity is required, and to remove adjacent elements O(n) time complexity is required.
• Auxiliary Space : O(1)
  Since no extra space is required, the space complexity is constant.

Method 3: The method involves the technique of Hashing, which takes O(n) time.

Approach: The time complexity in this method can be reduced but space complexity will take a toll.

This involves the use of Hashing where the numbers are marked in a HashMap, so that if the number is again encountered, then erase it from the array.

Step-by-step algorithm:

1. Use a hash set. HashSet stores only unique elements.
2. It is known that if two of the same elements are put into a HashSet, the HashSet stores only one element (all the duplicate element vanishes)
3. Traverse the array from start to end.
4. For every element, insert the element into the HashSet
5. Now Traverse the HashSet and put the elements in the HashSet in the array

Implementation:

C++

// C++ program to remove duplicates using Hashing
#include <bits/stdc++.h>
using namespace std;

int removeDups(vector<int>& vect)
{
    // Create a set from vector elements
    unordered_set<int> s(vect.begin(), vect.end());

    // Take elements from set and put back in
    // vect[]
    vect.assign(s.begin(), s.end());
}

// Driver code
int main()
{
    vector<int> vect{ 3, 5, 7, 2, 2, 5, 7, 7 };
    removeDups(vect);
    for (int i = 0; i < vect.size(); i++)
        cout << vect[i] << " ";
    return 0;
}

Java

// Java program to implement Naive approach to
// remove duplicates.
import java.util.*;

class GFG {

    static void removeDups(Vector<Integer> vect)
    {

        // Create a set from vector elements
        Set<Integer> set = new HashSet<Integer>(vect);

        // Take elements from set and put back in
        // vect[]
        vect.clear();
        vect.addAll(set);
    }

    // Driver code
    public static void main(String[] args)
    {
        Integer arr[] = { 3, 5, 7, 2, 2, 5, 7, 7 };
        Vector<Integer> vect
            = new Vector<Integer>(
                Arrays.asList(arr));
        removeDups(vect);
        for (int i = 0; i < vect.size(); i++) {
            System.out.print(vect.get(i) + " ");
        }
    }
}

/* This code contributed by PrinciRaj1992 */

C#

// C# program to implement Naive approach to
// remove duplicates.
using System;
using System.Collections.Generic;
using System.Linq;

class GFG {

    static List<int> removeDups(List<int> vect)
    {

        // Create a set from vector elements
        HashSet<int> set = new HashSet<int>(vect);

        // Take elements from set and put back in
        // vect[]
        vect.Clear();
        vect = set.ToList();
        return vect;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 3, 5, 7, 2, 2, 5, 7, 7 };
        List<int> vect = new List<int>(arr);
        vect = removeDups(vect);
        for (int i = 0; i < vect.Count; i++) {
            Console.Write(vect[i] + " ");
        }
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// JavaScript program to implement Naive approach to
// remove duplicates.

function removeDups(vect)
{
    // Create a set from vector elements
        letset = new Set(vect);

        // Take elements from set and put back in
        // vect[]
        vect=[];
        vect=Array.from(letset);
        return vect;
}

// Driver code
let vect=[3, 5, 7, 2, 2, 5, 7, 7 ];
vect=removeDups(vect);
for (let i = 0; i < vect.length; i++) {
    document.write(vect[i] + " ");
}


// This code is contributed by rag2127

</script>

Python3

# Python3 program to remove duplicates using Hashing
def removeDups():
    global vect

    # Create a set from vector elements
    s = set(vect)

    # Take elements from set and put back in
    # vect[]
    vect = s

# Driver code
vect = [3, 5, 7, 2, 2, 5, 7, 7]
removeDups()
for i in vect:
    print(i, end = " ")

# This code is contributed by shubhamsingh10

Output

2 7 5 3 

Complexity Analysis:

• Time Complexity: O(n).
  Since a single traversal is needed to enter all the elements in the hashmap, the time complexity is O(n).
• Auxiliary Space: O(n).
  For storing elements in HashSet or hashmap O(n) space complexity is needed.

Method 4: This focuses on small ranged values where the time complexity is O(n).

Approach:

This approach only works when the product of all distinct primes is fewer than 10^18 and all the numbers in the array should be prime. The property of primes of having no divisors except 1 or that number itself is used to arrive at the solution. As the array elements are removed from the array, they keep a value(product) which will contain the product of all distinct primes found previously in the array, so that if the element divides the product, they can be sure proved that the element has previously occurred in the array and hence the number will be rejected.

Step-by-step algorithm:

• Initially, keep a variable (p = 1).
• Traverse the array from start to end.
• While traversing, check whether p is divisible by the i-th element. If true, then erase that element.
• Else keep that element and update the product by multiplying that element with the product (p = p * arr[i])

Implementation:

C++

// Removes duplicates using multiplication of
// distinct primes in array
#include <bits/stdc++.h>
using namespace std;

int removeDups(vector<int>& vect)
{
    long long int prod = vect[0];
    int res_ind = 1;
    for (int i = 1; i < vect.size(); i++) {
        if (prod % vect[i] != 0) {
            vect[res_ind++] = vect[i];
            prod *= vect[i];
        }
    }
    vect.erase(vect.begin() + res_ind, vect.end());
}

// Driver code
int main()
{
    vector<int> vect{ 3, 5, 7, 2, 2, 5, 7, 7 };
    removeDups(vect);
    for (int i = 0; i < vect.size(); i++)
        cout << vect[i] << " ";
    return 0;
}

Java

// Removes duplicates using multiplication of
// distinct primes in array
import java.util.*;

class GFG {

    static int[] removeDups(int[] vect)
    {

        int prod = vect[0];
        int res_ind = 1;
        for (int i = 1; i < vect.length; i++) {
            if (prod % vect[i] != 0) {
                vect[res_ind++] = vect[i];
                prod *= vect[i];
            }
        }
        return Arrays.copyOf(vect, res_ind);
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] vect = { 3, 5, 7, 2, 2, 5, 7, 7 };
        vect = removeDups(vect);
        for (int i = 0; i < vect.length; i++)
            System.out.print(vect[i] + " ");
    }
}

// This code is contributed by 29AjayKumar

C#

// Removes duplicates using multiplication of
// distinct primes in array
using System;

class GFG {

    static int[] removeDups(int[] vect)
    {

        int prod = vect[0];
        int res_ind = 1;
        for (int i = 1; i < vect.Length; i++) {
            if (prod % vect[i] != 0) {
                vect[res_ind++] = vect[i];
                prod *= vect[i];
            }
        }
        int[] temp = new int[vect.Length - res_ind];
        Array.Copy(vect, 0, temp, 0, temp.Length);
        return temp;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int[] vect = { 3, 5, 7, 2, 2, 5, 7, 7 };
        vect = removeDups(vect);
        for (int i = 0; i < vect.Length; i++)
            Console.Write(vect[i] + " ");
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// Removes duplicates using multiplication of
// distinct primes in array

function removeDups(vect)
{
    let prod = vect[0];
        let res_ind = 1;
        for (let i = 1; i < vect.length; i++) {
            if (prod % vect[i] != 0) {
                vect[res_ind++] = vect[i];
                prod *= vect[i];
                vect=vect.slice(0,res_ind + 2);
            }
        }
        return vect;
}

// Driver code
let vect=[3, 5, 7, 2, 2, 5, 7, 7];
vect = removeDups(vect);
document.write(vect.join(" "));


// This code is contributed by rag2127
</script>

Python3

def remove_dups(vect):
    prod = vect[0]
    res_ind = 1

    for i in range(1, len(vect)):
        if prod % vect[i] != 0:
            vect[res_ind] = vect[i]
            res_ind += 1
            prod *= vect[i]

    del vect[res_ind:]


vect = [3, 5, 7, 2, 2, 5, 7, 7]
remove_dups(vect)

for num in vect:
    print(num, end=" ")

Output

3 5 7 2 

Complexity Analysis:

• Time Complexity: O(n).
  To traverse the array only once, the time required is O(n).
• Auxiliary Space: O(1).
  One variable p is needed, so the space complexity is constant.

Note: This solution will not work if there are any composites in the array.

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